Motion
Chapter 11

Standards
• Students will:
• SPS8.
Determine the relationship between force, mass and motion
• SPS8a Calculate velocity and acceleration
• SPS8c Relate falling objects to gravitational force
• SPS8d Explain difference in mass and weight

Observing Motion
• Motion - change in position in relation to a
reference point.

Measuring Motion: Distance
• Distance - how far an object moves on a path
• Displacement - how far between starting and ending points on a path

Measuring Motion: Speed
• Speed
– rate of motion
– distance traveled per unit time speed = distance time

Measuring Motion: Speed (cont’d)
• Instantaneous
Speed
– speed at a given instant
• Average Speed -
= total distance total time

Measuring Motion: Velocity
• Velocity
– speed in a given direction
– can change even when the speed is constant!

Calculating Velocity:
Distance/Speed/Time Triangle

Graphing Speed/Velocity
• X axis- usually independent variable (time)
• Y axis- usually dependent variable (distance)
• Slope of straight line= vertical change horizontal change

Graphing (cont’d)
• slope = velocity
• steeper slope = faster velocity
• straight line = constant velocity
• flat line = 0 velocity
(no motion)

Calculating Slope
Distance Vs Time
16
1. Choose two points on graph to calculate slope.
2. Calculate the vertical and horizontal change.
3. Divide the vertical change by the horizontal change.
14
12
10
8
6
4
2
.
.
0 1 2
Time (s)
3 4 5

Calculating Slope(cont’d)
Distance Vs Time
16
1. point 1: d= 6m t= 1s point 2: d= 12m t= 4s
2. vert ∆- 12m-6m= 6m horiz ∆- 4s-1s= 3s
14
12
10
8
6
.
Horizontal
Change
4
.
Vertical change
3. slope= 6m = 2m/s
3s
2
0 1 2
Time (s)
3 4 5

Practice Problem
A
B
• Who started out faster?
– A (steeper slope)
• Who had a constant speed?
– A
• Describe B from 10-20 min.
– B stopped moving
• Find their average speeds.
– A = (2400m) ÷ (30min)
A = 80 m/min
– B = (1200m) ÷ (30min)
B = 40 m/min

Measuring Motion: Acceleration
• Acceleration
– the rate of change of velocity
– change in speed or direction
• Centripetal acceleration
– circular motion (even if speed is constant, direction is always changing) ex. Moon accelerates around Earth

Measuring Motion: Acceleration
• Positive acceleration
-“speeding up”
• Negative acceleration
-“slowing down”

Calculating Acceleration
• a= v f – v i t
• a- acceleration
• v ffinal velocity
• v iinitial velocity
• t- time
• a= ∆ v t

Calculating Acceleration (cont’d)
1. List given, then
unknown values.
2. Write equation for acceleration.
3. Insert known
values into equation and solve.
t
V f -V i a t

Practice Problem 1
A flowerpot falls off a second-story windowsill.
The flowerpot starts from rest and hits Mr.
Mertz 1.5s later with a velocity of 14.7m/s.
Find the average acceleration of the flowerpot.
Given: Remember: Solve: t= 1.5s a= v f – v i a= 14.7m/s-0m/s
Vi= 0m/s t 1.5s
Vf= 14.7m/s a= 14.7m/s a= ? 1.5s
a= 9.8m/s2

Practice Problem 2
Joseph’s car accelerates at an average rate of
2.6m/s2. How long will it take his car to speed up from 24.6m/s to 26.8m/s2?
Given: Remember: Solve: a= 2.6m/s2 t= (vf-vi) ÷ a vf= 26.8m/s2 vf-vi t= 26.8m/s2-24.6m/s2
Vi= 24.6m/s2 a t 2.6m/s2 t= ? t= 2.2m/s
2.6m/s2 t= 0.85s

Practice Problem 3
A cyclist travels at a constant velocity of 4.5m/s westward and then speeds up with a steady acceleration of 2.3m/s2. Calculate the cyclist’s speed after accelerating for 5.0s.
Given: Remember: Solve: vi= 4.5m/s vf= vi + a x t vf= ? vf-vi vf= 4.5m/s + (2.3m/s2 x 5.0s) a= 2.3m/s2 a t vf= 4.5m/s + 11.5m/s t= 5.0s vf= 16m/s

400
300
Graphing Acceleration
Distance/Time
On Distance-Time graph:
• Acceleration is shown as a curved line
200
100
0
0 5 10
Time (s)
15 20

3
2
1
0
0 2
Graphing Acceleration
Speed/Time
On a Speed-Time graph:
• Slope of straight line= acceleration
• Positive slopespeeding up
• Negative slope- slowing down
• Flat line- constant velocity
(no acceleration)
4
Time (s)
6 8 10

Newton’s Laws of Motion
• 2 nd Law of Motion:
The acceleration of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass.
Force = mass x acceleration or or
F= ma

Force
Force - push or pull that one body exerts on another

Fundamental Forces
• 4 types:
1. gravity
2. electromagnetic
3. weak nuclear
4.
strong nuclear
• Vary in strength
• Act through contact or at a distance

Forces
• Force Pairs : forces moving in opposite directions
• Balanced forces : do not
move; push equally on each other
• Unbalanced forces : acceleration (movement) in the direction of larger force

Force Pair Examples

Friction
Friction : force that opposes motion between 2 surfaces
• Static Friction : nonmoving surfaces
• Kinetic Friction : moving surfaces- sliding or rolling
(sliding friction is greater than rolling friction)

Friction Facts
• Necessary for all motion
• Rougher surfaces create greater friction
• Greater mass creates greater friction
• Lubricants reduce friction

New ton’s First and Second La ws
Chapter 12

Newton’s Laws of Motion
• 1 st Law of Motion: Law of Inertia
• An object at rest will remain at rest;
• An object in motion will continue moving at a constant velocity unless acted upon by a net force.

Inertia
• Objects move only when net force is applied.
• Objects maintain state of motion.
• Inertia is related to mass.
(small mass can be accelerated by small force large mass can be accelerated by large force)

Newton’s Laws of Motion
• 2 nd Law of Motion:
The acceleration of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass.
Force = mass x acceleration or or
F= ma

Newton’s Second Law
• For equal forces, large masses accelerate less
• Force is measured in newtons (N)
• 1N= 1kg x 1m/s2
F m a

Practice Problem 1
Zoo keepers lift a stretcher that holds a sedate lion. The lion’s mass is 175 kg and the upward acceleration of the lion and stretcher is 0.657m/s2.
What force is needed to produce acceleration of the lion and stretcher?
Given: Remember: Solve: m= 175 kg F = m x a
F a= 0.657m/s2 F = 175 kg x 0.657m/s2
F
F= ? m a F = 11.49 N

Practice Problem 2
A baseball accelerates downward at 9.8 m/s2. If gravity is the only force acting on the baseball and is
1.4N, what is the baseball’s mass?
Given: Remember: Solve:
F= 1.4 kg/m/s2 m = F m = 1.4 N
9.8 m/s2 m = .14 kg

Weight and Mass
• Weight - measure of gravity on an object
• Not equal to mass (constant everywhere)
• Measured in Newtons
weight = mass x free-fall acceleration
(9.8m/s2) w free-fall m g acceleration

Gravity
• Force of attraction between 2 objects in the universe
• Increases as:
- mass increases
- distance decreases
• Affects all matter

Gravity Quiz 1
Who experiences more gravity - the astronaut or the politician?
More distance
Less distance

Gravity Quiz 2
Which exerts more gravity, your hand or your pencil ?

Gravity Quiz 3
• Would you weigh more on Earth or Jupiter?
(Hint: Which planet has the greater mass?)

Free Fall Acceleration
• Occurs when Earth’s gravity is only force acting on an object
• In absence of air
resistance, all objects accelerate at same rate
• g = 9.8 m/s2
• g = 9.8 m/s 2 g = 9.8 m/s 2 g = 9.8 m/s 2 g =
9.8 m/s 2 g = 9 .8 m/s 2

Air Resistance
• Force of air on a moving object which
opposes its motion
• Aka fluid friction or drag
• Depends on objects:
- speed
- surface area
- shape
- density

Air Resistance (cont’d)
• Terminal velocity= maximum velocity reached by a falling object
• Reached when…
F gravity = F air resistance
(no net force)

Projectile Motion
Projectile-
• Any object thrown in air
• Only acted on by gravity
• Follows parabolic pathtrajectory
• Has horizontal and
V ox
V oy vertical velocities

Projectile Motion (cont’d)
• Horizontal Velocity
– depends on inertia
– remains constant
• Vertical Velocity
– depends on gravity
– accelerates downward at
9.8 m/s 2

Projectile Motion Quiz
• A moving truck launches a ball vertically
(relative to the truck). If the truck maintains a constant horizontal velocity after the launch, where will the ball land (ignore air resistance)?
A) In front of the truck
B) Behind the truck
C) In the truck
Answer: C because horizontal and vertical velocities are independent of each other

Newton’s Laws of Motion
• 3 rd Law of Motion:
When one object exerts a force on a second object, the second object exerts an equal but
opposite force on the first.

Newton’s Third Law
• Forces always occur in pairs
• Forces in a pair do not act on the same object
• Equal forces don’t always have equal effects

Newton’s Third Law
Problem: How can a horse pull a cart if the cart is pulling with equal force back on the horse?

Newton’s Third Law
Answer:
1. Forces are equal and opposite but are acting on different objects
2. Forces are not balanced
3. The movement of the horse depend on the forces working on the horse.

Momentum
Momentum-
• quantity of motion
• = mass x velocity
• units: kg x m/s p m v

Practice Problem 1
Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s.
Given Remember Solve m = 280 kg p = mv p v = 3.2 m/s p =(280kg)(3.2m/s) m vv v
P = ? P = 896 kg  m/s

Practice Problem 2
The momentum of a second bumper car is 675
kg·m/s. What is its velocity if its total mass is
300 kg?
Given Remember Solve m = 300kg v =(675kg·m/s)÷(300kg) m v v = ?
v = 2.25m/s

Conservation of Momentum
Law of Conservation of Momentum:
• Total momentum of two or more objects before a collision is the same as it was before the collision (momentum is conserved).

Jet Car Challenge
CHALLENGE:
Construct a car that will travel a least 3 meters using only the following materials:
•
• tape
• 4 plastic lids
• 2 plastic straws
•
•
• 2 skewers
1 balloon
1 cardboard base
Use your knowledge of Newton’s Laws to make this thing go!